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Let $\alpha = 7+4\sqrt{3}$. Consider the series $$ \sum_{n\geq 0}\cos\left(\frac{\pi}{4}\alpha^n\right) $$

I'm having a doubt. On one hand, I've plotted (I think) the evolution of the partial sum $\left(S_n\right)$ and I've obtained:

So I've assumed that it was divergent. However, it can be easily shown that $$ \frac{1}{2}\left(\alpha^n+\frac{1}{\alpha^n}\right) = 2p_n+1 $$ where $p_n$ is a sequence positive integers. As a result $$ \cos\left(\frac{\pi}{4}\alpha^n\right) = \cos\left(\frac{\pi}{4}\left[4p_n+2-\frac{1}{\alpha^n}\right]\right) = \cos\left(\frac{\pi}{2}+\pi p_n -\frac{\pi}{4\alpha^n}\right)=\left(-1\right)^{p_n}\sin\left(\frac{\pi}{4\alpha^n}\right) $$ Noting $\alpha \approx 14 > 1$, I've concluded $$ \left|\cos\left(\frac{\pi}{4}\alpha^n\right)\right| \underset{(+\infty)}{\sim}\frac{\pi}{4}\frac{1}{\alpha^n} $$ which is then is the general term for a convergent series.

So does it converge or diverge? Could someone point me in the right direction?

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  • $\begingroup$ You almost did it. Note that the absolute value decreases to $0$. $\endgroup$ Commented Jul 6, 2022 at 21:29
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    $\begingroup$ The word is "series" (that's both singular and plural), not "serie". $\endgroup$
    – jjagmath
    Commented Jul 6, 2022 at 21:30
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    $\begingroup$ it surely converges since it is absolutely convergent as you've shown $\endgroup$
    – Exodd
    Commented Jul 6, 2022 at 21:31
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    $\begingroup$ When computing the terms numerically, it is possible there is some rounding error in computing $\alpha^n$, which can result in the calculated terms not approaching zero. $\endgroup$
    – JimmyK4542
    Commented Jul 6, 2022 at 21:31

1 Answer 1

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Here's a plot with regular floating point operations:

flotaing point

and here's a plot with high precision operations:

enter image description here

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    $\begingroup$ What did you use to produce these? $\endgroup$ Commented Jul 6, 2022 at 21:47
  • $\begingroup$ I used Mathematica $\endgroup$
    – jjagmath
    Commented Jul 6, 2022 at 21:50
  • $\begingroup$ Thanks for your clear answer ! I've tried it on Wolfram Alpha, it told me that the series diverge. Do you have an explanation ? $\endgroup$
    – Atmos
    Commented Jul 7, 2022 at 7:27
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    $\begingroup$ @Atmos Yes, the explanation is very simple: Wolfram Alpha have bugs. $\endgroup$
    – jjagmath
    Commented Jul 7, 2022 at 10:42

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